Design of Rectangular Combined Footing<Program Created by: Engr. Jeremy E Caballes>
P1 DL = 25 kN P2 DL = 25 kN
P1 LL = 31 kN P2 LL = 31 kN
M1 DL = 0.1 kN-m M2 DL = -0.1 kN-m
E = 200,000.00 MPa M1 LL = 0.3 kN-m M2 LL = -0.3 kN-m
f'c = 20.70 MPa
fy = 275.00 MPa
Wconc = 23.50 kN per cu m
Wsoil = 16.50 kN per cu m
qa = 144.00 kPa
OR
qe = kPa
D = 1.5 m
Assume T = 300 mm
0.3
Remarks: SAFE!
Distance from the Column Spacing = 2.1 m
Property Line = m
C1y = 400 mm C2y = 400 mm
0.4 0.4
Main bar Ø = 16 mm
0.016
Concrete cover = 75 mm
0.075
Temperature bar Ø = 12 mm
C1x = 400 mm C2x = 400 mm
Optional W = m 0.4 0.4
Computed dimensions: W = 0.39 m and L = 2.5 m
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
Solve for L to give a uniform pressure:
P1 = P1 DL + P2 LL
P1 = 56.000 kN
P2 = P2 DL + P2 LL
P2 = 56.000 kN
M1 = M1 DL + M2 LL
M1 = 0.400 kN-m
M2 = M2 DL + M2 LL
M2 = -0.400 kN-m x-factor = 2 1.7
X1 = 0.200 m
X2 = 2.300 m 0.5
P = P1 + P2 0.000 0.000 -1.5
P = 112.000 kN
P(Xc) = P1(X1) + P2(X2) + M1 +M2 0.000 0.400 -1.110 1.100 -1.100 4.200 2.100
Xc = 1.250 m ≤ X2, OK! 5.000 5.000 -1.110 1.100 -1.100 5.000 2.100
L = 2Xc 5.000 5.000 -1.890 0.500 -1.900 5.000 1.100
L = 2.500 m say 2.500 m 0.000 0.000 -1.890 0.500 -1.900 4.200 1.100
Determine the width, W: 0.000 0.000 -1.110 1.100 -1.100 4.200 2.100
qe = qa - qs - qc
qs = Wsoil x (D - T)
qs = 19.800 kPa 0.000 4.5
qc = Wconc x T 0.000 -6.250
qc = 7.050 kPa
Therefore: qe = 117.150 kPa 0.000 4.5 1.234 3.766 5.000
Thus: A = P/qe 0.000 -6.250 1.234 3.766 5.000
A = 0.956 sq m
W = A/L Pres. fac = 0.006
W = 0.390 m say 0.390 m 0.000 -5.000 0.000 -1.100
Therefore Use: L = 2.500 m 5.000 -5.000 0.800 -1.100
W = 0.390 m 5.000 -6.000 0.800 -1.900
Assume: A = LW 0.000 -6.000 0.000 -1.900
A = 0.975 sq m 0.000 -5.000 0.000 -1.100
Design loads:
Pu1 = 1.4(P1 DL) + 1.7(P1 LL)
Pu1 = 87.700 kN
Pu2 = 1.4(P2 DL) + 1.7(P2 LL)
Pu2 = 87.700 kN
Pu = Pu1 + Pu2
Pu = 175.400 kN
Mu1 = 1.4(M1 DL) + 1.7(M1 LL)
Mu1 = 0.650 kN-m
Mu2 = 1.4(M2 DL) + 1.7(M2 LL)
Mu2 = -0.650 kN-m
Mu = Mu1 + Mu2
Mu = 0.000 kN-m
Assume: qu = P/A qu = 179.897 kPa
qu = 179.897 kPa w1 = 70.16 kN/m kN/m
qu2 = 179.897 kPa w2 = 70.16 kN/m kN/m Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
2222
Critical Section for One-Way or Direct Shear
70
.16
kN
/m
70
.16
kN
/m
Check for One-Way or Direct Shear:
d = T - 0.5Ø - cover d = 217 mm
(a) Exterior Column.: 0.000 4.500 1.017 3.983 5.000
Xa = X1-C1x/2-d Xa = 0.000 m 0.000 -6.250 1.017 3.983 5.000
Xb = X1+C1x/2+d Xb = 0.617 m
qua = 179.897 kPa
qub = 179.897 kPa
Vua = 0.000 kN 0 0.000 -0.883 3.983 -0.883
Vub = -44.411 kN 44.41128 1.017 -0.883 5.000 -0.883
(b) Interior Column: 1.017 -2.117 5.000 -2.117
Xc = X2-C2x/2-d Xc = 1.883 m 0.000 -2.117 3.983 -2.117
Xd = X2+C2x/2+d Xd = 2.500 m 0.000 -0.883 3.983 -0.883
quc = 179.897 kPa
qud = 179.897 kPa V Scale = 0.022805017
Vuc = 44.411 kN 0.000 -5.000
Vud = 0.000 kN 0.000 -5.000
0.4 -7
Governing Vu = 44.411 kN 1.234 -6.013
2.500 -5.000
ØVc = Ø(1/6)√(f'c)bd 3.766 -3.9872
ØVc = 54.548 kN SAFE! 4.6 -5.9872
5.000 -5
Check for Two-Way or Punching Shear: 5.000 -5.000
(a) Exterior Column.:
Xa = X1-C1x/2-d/2 Xa = 0.000 m
Xb = X1+C1x/2+d/2 Xb = 0.509 m
qua = 179.897 kPa
qub = 179.897 kPa
x1 = 0.509 m
y1 = 0.617 m
Vu = Pu1 - 0.50(qua+qub)x1y1
Vu = 31.258 kN ßc = 1.000
bo = 1.634 m 4 4
ØVc = Ø(1/3)√(f'c)bod (1/3) 0.3333333333
ØVc = 457.083 kN SAFE!
(b) Interior Column:
Xc = X2-C2x/2-d/2 Xc = 1.992 m
Xd = X2+C2x/2+d/2 Xd = 2.500 m
quc = 179.897 kPa
qud = 179.897 kPa
x2 = 0.508 m 2.251
y2 = 0.617 m
Vu = Pu2 - 0.50(quc+qud)x2y2
Vu = 31.258 kN ßc = 1.000
bo = 1.634 m 4 4
ØVc = Ø(1/3)√(f'c)bod (1/3) 0.3333333333
ØVc = 457.083 kN SAFE!
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
3
222
43
2
Critical Section for Two-Way or Punching Shear
70
.16
kN
/m
70
.16
kN
/m
Design of Reinforcement:
(a) Long direction. (d = T - 0.5Ø - cover) d = 217.000 mm
0.000 4.500 0.800 2.500 4.200 5.000
Xa = X1 - d/2 Xa = 0.000 m 0.000 -6.250 0.800 2.500 4.200 5.000
Xb = X1 + d/2 Xb = 0.400 m
Xc = X[V=0] Xc = 1.250 m 0 0.01
Xd = X2 - d/2 Xd = 2.100 m 0 6 0.01
Xe = X2 + d/2 Xe = 2.500 m 350.8
-438.5
Mua = 0.000 kN-m Mua = 0.000E+0 N-mm
Mub = -11.277 kN-m Mub = -11.277E+6 N-mm
Muc = -36.623 kN-m Muc = -36.623E+6 N-mm
Mud = -11.277 kN-m Mud = -11.277E+6 N-mm
Mue = 0.000 kN-m Mue = -63.949E-9 N-mm
4√f`cor
1.4
fy fy
= 0.004 = 0.0050909091
0.00509
Ab = πd² / 4
Ab = 201.062 sq mm
Mu = 0.000 -11.277 -36.623 -11.277 0.000
Section a b c d e
Mu = 0.000E+0 11.277E+6 36.623E+6 11.277E+6 63.949E-9
b = S = 390.000 390.000 390.000 390.000 390.000
Rn = 0.00000 0.68230 2.21575 0.68230 0.00000
act p = 0.00000 0.00253 0.00864 0.00253 0.00000
Use: p = 0.00509 0.00509 0.00864 0.00509 0.00509
As = 430.844 430.844 731.267 430.844 430.844
n = 3 3 4 3 3
Soc = 112 112 74 112 112
Scl = 96.000 96.000 58.000 96.000 96.000
Location = Bottom Top Top Top Bottom
112.000 112.000
112.000 74.000 112.000
Min Spacing: Bottom Top n
a-b 112.000 3
c 74.000 4
d-e 112.000 3
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
ρmin = ρmin =
Use ρmin =
2 222
Critical Section for Bending (Long Direction)
70
.16
kN
/m
70
.16
kN
/m
(a) Short direction. (d = T - 0.5Ø - cover) d = 217.000 mm 0.000 4.500 1.126 3.875 5.000
d factor = 0.75 0.217 m 0.000 -6.250 1.126 3.875 5.000
b1 = C1x + 0.75d b2 = C2x + 0.75d 0.000 -1.110 3.875 -1.110
b1 = 562.750 mm b2 = 562.750 mm 1.126 -1.110 5.000 -1.110
d1 = W/2 - C1y/2 d2 = W/2 - C2y/2 1.126 -1.100 5.000 -1.100
d1 = -5.000 mm d2 = -5.000 mm 0.000 -1.100 3.875 -1.100
0.000 -1.110 3.875 -1.110
Xa = 0.000 m
Xb = 0.563 m
Xc = 1.937 m
Xd = 2.500 m
M1 = qu(b1)(d1²/2) M2 = qu(b2)(d2²/2)
M1 = 1,265 N-mm M2 = 1,265 N-mm
Mu = 1,265 1,265
Section b1 x d1 b2 x d2
Mu = 1.265E+3 1.265E+3
b = W = 562.750 562.750
Rn = 0.00005 0.00005
act p = 0.00000 0.00000
Use: p = 0.00509 0.00509
As = 621.685 621.685
n = 4 4
Soc = 137 137
Scl = 121.000 121.000
Location = Bottom Bottom
Temperature/Shrinkage Reinforcement: Ø = 12 mm
Ast = 0.002bh
Ast = 600.000 sq mm
n = 5.305 say 6
Soc = 200.000 mm
Scl = 188.000 mm > 25 mm OK!
n = 0 to 0 m
n = 5 0.563 m to 1.937 m
n = 2 2.5 m to 2.5 m
n = 2 W
Programmed by : Engr. Jeremy E. Caballes, 17 April 2004, Revised, 11 Nov 2005
1435
2222
Critical Section for Bending (Short Direction)
qu
= 1
79
.89
7 k
Pa
qu
= 1
79
.89
7 k
Pa
f'c = 20.7 MPa Increase = -3.5
fy = 275 MPa 0.000 0.600 0.000 2.000 4.200 2.000
Wconc = 23.5 kN per cu m 5.000 0.600 0.800 2.000 5.000 2.000
Wsoil = 16.5 kN per cu m 5.000 0.000 0.800 0.600 5.000 0.600
qa = 144 kPa 0.000 0.000 0.000 0.600 4.200 0.600
T = 300 mm 0.000 0.600 0.000 2.000 4.200 2.000
Edge Dist. = 0.2 m
Col Dist. = 2.1 m
Edge Dist. = 0.2 m 0.000 -3.110 0.000 -3.100 4.200 -3.100
Ext. Col: 400 mm x 400 mm 5.000 -3.110 0.800 -3.100 5.000 -3.100
Int. Col: 400 mm x 400 mm 5.000 -3.890 0.800 -3.900 5.000 -3.900
Main bar Ø = 16 mm 0.000 -3.890 0.000 -3.900 4.200 -3.900
Concrete cover = 75 mm 0.000 -3.110 0.000 -3.100 4.200 -3.100
Temperature bar Ø = 12 mm
L = 2.5 m 0.166 -3.110 1.1255 3.8745 5 5.000 0
W = 0.39 m 0.166 -6.500 1.1255 3.8745 5 5.000 -3.110
D = 1.5 m
d = 217 mm Edge Dist. = 0.166
0.166 -3.276 -1 -3.110 0.166 -6.5 -0.500 -3.276
4-16 @ 137 mm 4.834 -3.276 -1 -3.890 0.166 -6.5 -0.500 -3.724
5-12 @ 200 mm 4.834 -3.724 1.1255 -6.5
4-16 @ 137 mm 0.166 -3.724 6.000 -3.276 3.8745 -6.5
0.166 -3.110 6.000 -3.724 5 -6.5
Bottom 3-16 @ 112 mm 4.834 -6.5
Top 4-16 @ 74 mm 0.166 -3.724 0.166 0.166
Bottom 3-16 @ 112 mm 0.166 -6.5 4.834 0.166 0.000 -1
0.400 -1
4.834 -3.724 0.000 0.434 4.6 -1
4.834 -6.500 5.000 0.434 5.000 -1
Optional W = m -1 -3.110 -1 3.000 0.4 -1.000
Computed dimenComputed dimension 0.000 -3.110 6.000 3.000 0.4 -3.500
-1 -3.890 -1 0.000 4.6 -1.000
0 -3.890 6.000 0.000 4.6 -3.500
-0.500 -3.276 -1 3.000 -0.5 0.600
6.000 -3.276 -1 0.000 -0.5 0.166
-0.500 -3.724 6.000 0.600 -0.5 0.600 -0.5 0.166
6.000 -3.724 6.000 0.000 6.000 0.600 0.000 0.166
-0.500 -3.724 0.000 -0.5 -0.5 0.166
0 -0.5
6.000 -3.724 5.000 -0.5 0.000 0.166
Reinforcement Detail of Combined Rectangular FootingBy: Engr. Jeremy E. Caballes
L = 2.5 m
W =
0.3
9 m
T =
30
0
mmD
= 1
.5 m
Col Dist. = 2.1 m
Ext. Col: 400 mm x 400 mm
4-16 @ 137 mm 4-16 @ 137 mm
5-12 @ 200 mm
3-1
6 @
11
2 m
m
4-1
6 @
74
mm
Bo
tto
m
To
p
Edge Dist. = 0.2 m
3-1
6 @
11
2 m
m
Bo
tto
m
f'c = 20.7 MPa
fy = 275 MPa
Wconc = 23.5 kN per cu m
Wsoil = 16.5 kN per cu m
qa = 144 kPa
d =
21
7 m
m
Main bar Ø = 16 mm
Concrete cover = 75 mm
Temperature bar Ø = 12 mm
Int. Col: 400 mm x 400 mm
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